Kepler Orbits and the Laplace-Runge-Lenz Vector

Derivation

Kepler Orbits and the Laplace-Runge-Lenz Vector

D-131 Home PU-201 Threads force · symmetry · energy Depends on The Central-Force Orbit Equation, Effective Potential and Radial Reduction

Statement

For a particle of (reduced) mass m moving under the attractive inverse-square central force F(r) = −k/r² r̂ (k > 0), the bound and unbound orbits are conic sections r(θ) = p/(1 + e cosθ) with semi-latus rectum p = L²/(mk) and eccentricity e = √(1 + 2EL²/mk²); and the Laplace–Runge–Lenz vector A = p × L − mk r̂ is a constant of the motion whose conservation forces the orbit to close and its major axis to stay fixed (no precession).

Why it matters

The inverse-square law is one of only two central forces (the other being the isotropic harmonic oscillator) whose bound orbits are exactly closed. Every other power law produces a rosette that slowly precesses. The special closure is not an accident of solving the differential equation — it is enforced by an extra vector conservation law that has no analogue in a generic central problem.

Historically this is why Newtonian gravity predicts stationary Keplerian ellipses, and why the observed 43″/century precession of Mercury — a tiny breaking of the LRL symmetry — became the sharpest classical test of general relativity. The same vector, promoted to a quantum operator, reproduces the exact n⁻² Bohr spectrum of hydrogen and explains its hidden SO(4) degeneracy.

Assumptions

Force is exactly inverse-square and central.Any deviation (oblateness, a 1/r³ correction, a third body) breaks the closure; the orbit becomes a precessing rosette and A is no longer conserved. Motion reduces to a one-body problem of mass m.For two comparable masses use the reduced mass μ = m₁m₂/(m₁+m₂) about the centre of mass; dropping this misplaces the focus and mis-scales the period. Angular momentum L ≠ 0.With L = 0 the "orbit" is a radial free-fall line, p = 0, and the conic form degenerates.

Non-relativistic, point masses, no radiation.Special-relativistic velocity dependence or gravitational/electromagnetic radiation reaction adds effective higher-order terms; A then drifts and the ellipse precesses or spirals.

Derivation

F(r) = −k/r², u ≡ 1/r, d²u/dθ² + u = −m/(L²u²) · F(1/u)

Import the central-force orbit (Binet) equation, valid for any central force at fixed L. A

−m/(L²u²) · (−k u²) = mk/L² ⇒ d²u/dθ² + u = mk/L²

Substitute F(1/u) = −ku²; the u² cancels, leaving a linear ODE with constant forcing. This cancellation is what makes inverse-square special. B

u(θ) = mk/L² + C cos(θ − θ₀)

General solution = particular constant + homogeneous simple-harmonic solution in θ (the operator d²/dθ² + 1). B

choose θ₀ = 0, r = 1/u = (L²/mk) / (1 + (CL²/mk) cosθ)

Orient the polar axis along the direction of closest approach and invert u. Pure choice of axis, no loss of generality. A

p ≡ L²/mk, e ≡ CL²/mk ⇒ r = p/(1 + e cosθ)

Identify the standard polar equation of a conic with focus at the force centre (semi-latus rectum p, eccentricity e). A

E = ½mṙ² + L²/(2mr²) − k/r, ṙ = (L/m) du/dθ ×(−r²·u...)

Bring in the conserved energy from the effective-potential result; use ṙ = −(L/m) du/dθ to express everything through u(θ). B

E = L²/(2m) [ (du/dθ)² + u² ] − ku = L²/(2m)(C² − m²k²/L⁴)

Insert u = mk/L² + Ccosθ; the θ-dependent terms cancel identically (they must, since E is constant), leaving E in terms of the amplitude C. C

C² = m²k²/L⁴ (1 + 2EL²/mk²) ⇒ e = CL²/mk = √(1 + 2EL²/mk²)

Solve for the amplitude and convert to eccentricity via the definition of step 5. This ties the geometric e to the dynamical invariants E, L. B

A ≡ p × L − mk r̂, with p = mv⃗, L = r × p

Define the Laplace–Runge–Lenz vector. We now show it is conserved directly from the equation of motion. C

dp/dt = F = −(k/r²) r̂, dL/dt = 0 ⇒ d(p×L)/dt = F × L

Angular momentum is conserved for any central force, so only p varies inside the cross product. B

F×L = −(k/r²) r̂ × (m r²θ̇ ẑ) = mkθ̇ θ̂ = mk dr̂/dt

Write L = mr²θ̇ ẑ and use r̂×ẑ = −θ̂, then dr̂/dt = θ̇ θ̂. The two rates are identical. C

dA/dt = d(p×L)/dt − mk dr̂/dt = mk dr̂/dt − mk dr̂/dt = 0

The two terms cancel exactly — only the inverse-square law makes F×L a total time derivative of mkr̂. Hence A is conserved. C

A·r = (p×L)·r − mkr = L·(r×p) − mkr = L² − mkr

Use the scalar-triple-product identity (p×L)·r = L·(r×p) = L·L. C

Arcosθ = L² − mkr ⇒ r = (L²/mk)/(1 + (A/mk)cosθ)

Write A·r = Arcosθ with θ the angle from A, and solve for r. This reproduces the conic of step 5 with no integration — and identifies e = A/mk, A pointing toward perihelion. C

A² = m²k² + 2mEL² ⇒ e = A/mk = √(1 + 2EL²/mk²)

Expand A² = (p×L)² − 2mk r̂·(p×L) + m²k² using (p×L)² = p²L² and E = p²/2m − k/r; the two eccentricities agree. C

Result

r(θ) = p/(1 + e cosθ), p = L²/mk, e = √(1 + 2EL²/mk²); A = p×L − mkr̂ = const, |A| = mke

Reading. The orbit is a conic with the force centre at one focus. Energy sets the shape: E < 0 gives an ellipse (e < 1, semi-major axis a = −k/2E), E = 0 a parabola (e = 1), E > 0 a hyperbola (e > 1). The vector A lies in the orbital plane, points from focus to perihelion, and has fixed length mke; its constancy is exactly the statement that the perihelion never moves, so a bound orbit closes on itself after one revolution.

Units check. [L²/mk] = (kg m² s⁻¹)² / (kg · kg m³ s⁻²) = m, so p is a length. In 2EL²/mk²: (J)(kg m² s⁻¹)²/[kg (kg m³ s⁻²)²] is dimensionless, so e is a pure number. [A] = [p][L] = (kg m s⁻¹)(kg m² s⁻¹) = kg² m³ s⁻² = [mk], consistent, and A/mk = e is dimensionless.

Limiting cases

Circular orbit (e = 0): requires E = −mk²/2L², the minimum of the effective potential; here A = 0 and no direction is singled out.
Marginal escape (E → 0⁻): e → 1⁻, a → ∞, the ellipse degenerates into a parabola of the same p.
Strong deflection (E ≫ 0 or L → 0): e ≫ 1, a nearly straight hyperbola; the asymptote angle satisfies cosθ_∞ = −1/e.
Repulsive Coulomb (k < 0, e.g. Rutherford): only E > 0 hyperbolae exist, with the force centre at the far focus; the same A derivation holds with the sign of k flipped.

Breaks when

The potential is not pure 1/r. A quadrupole term (planetary oblateness, J₂), an added 1/r³ piece, or third-body tugs make dA/dt ≠ 0; the ellipse precesses at a rate proportional to the perturbation, and the closed-orbit theorem no longer applies.
Relativistic dynamics. In GR the effective potential gains a −GML²/c²r³ term, giving the perihelion advance Δφ = 6πGM/[c²a(1−e²)] per orbit — the Mercury result. The Newtonian LRL vector is only approximately conserved.
Radiation / dissipation. An accelerating charge (or an inspiralling binary emitting gravitational waves) loses energy; E and L decay, a and e shrink, and the orbit spirals inward rather than closing.

Failure modes

Focus at the centre of the ellipse. The force centre sits at a focus, not the geometric centre; perihelion is a(1−e), aphelion a(1+e).
Using total mass instead of reduced mass. For comparable masses forgetting μ mis-scales L²/mk and the period.
Dropping the u² cancellation. Writing the orbit equation with F ∝ 1/r² but not converting to F(1/u) = −ku² leaves a nonlinear ODE and hides why the orbit is a clean conic.
Confusing the two "p"s. Here p is linear momentum in A, while p = L²/mk is the semi-latus rectum — different objects sharing a letter.
Claiming every central force gives closed orbits. Only 1/r² and r (Bertrand's theorem) do; students over-generalise the conic result.
Sign of k. Taking k < 0 (repulsion) but still expecting a bound ellipse; repulsion admits only hyperbolae.

Discussion

The deep reason the inverse-square orbit closes is symmetry. Every central force conserves energy and the three components of L (rotational SO(3) symmetry), and that alone confines the motion to a plane and to an annulus between turning points — but it permits the apsides to precess. The inverse-square force possesses an additional hidden symmetry whose Noether charge is A. With A conserved as well, the perihelion direction is frozen, and a bound orbit must retrace itself exactly.

For bound states this enlarged symmetry is SO(4): the six conserved quantities are the three components of L and three suitably scaled components of A (namely A/√(−2mE)), which together close a Lie algebra isomorphic to that of rotations in four dimensions. For scattering states (E > 0) the group is SO(3,1), and for E = 0 the Euclidean group E(3). This is precisely the "accidental" degeneracy that in quantum mechanics makes the hydrogen energy depend only on the principal quantum number n and not on ℓ.

Because A and L together over-determine the orbit, the Kepler problem is maximally superintegrable: it has 2N−1 = 5 independent constants of motion in three dimensions, one more than needed for Liouville integrability. That extra constraint is what pins the orbit to a fixed closed curve rather than letting it fill a torus densely. Perturbations that break the hidden symmetry — a J₂ bulge, general-relativistic corrections, or the diamagnetic term in a strong magnetic field — lift the degeneracy first, showing up as slow apsidal precession long before they change the orbit's size; the precession rate is a direct measurement of how strongly the 1/r symmetry is violated.

Common misconceptions. The LRL vector is often thought to be a peculiarity of gravity; it is a property of any exact 1/r potential, attractive or repulsive, and appears identically in Coulomb scattering. It is also not an "extra force" — it is a bookkeeping vector built from the state (r, p) that happens to stay constant. And a precessing planetary orbit does not mean the LRL vector is wrong; it means the real potential is not exactly 1/r.

Worked examples

Low Earth satellite: find e and a from E, L. m = 500 kg at r = 7.00×10⁶ m, tangential speed v = 8.00 km/s; GM_⊕ = 3.986×10¹⁴ m³s⁻², so k = GM_⊕m = 1.993×10¹⁷.

Set up invariants from the state. A

L = mvr = 500 × 8.00×10³ × 7.00×10⁶ = 2.80×10¹³ kg m² s⁻¹

Purely tangential velocity, so L = mvr directly. A

E = ½mv² − k/r = 1.60×10¹⁰ − 2.847×10¹⁰ = −1.247×10¹⁰ J

Bound (E < 0). A

2EL²/mk² = 2(−1.247×10¹⁰)(7.84×10²⁶) / [500(1.993×10¹⁷)²] = −0.984

Plug into the eccentricity formula's radicand. B

e = √(1 − 0.984) = √0.0157 = 0.126; a = −k/2E = 1.993×10¹⁷/(2×1.247×10¹⁰) = 7.99×10⁶ m

Evaluate e and semi-major axis. B

e ≈ 0.13, a ≈ 7.99×10³ km

Reading. A mildly elliptical LEO: perihelion a(1−e) ≈ 6.99×10³ km, aphelion a(1+e) ≈ 9.00×10³ km. Cross-check: p = L²/mk = 7.87×10⁶ m equals a(1−e²) = 7.87×10⁶ m. Units check. a in metres, e dimensionless.

Interstellar flyby: hyperbolic e and perihelion. A body approaches the Sun with speed at infinity v_∞ = 26.0 km/s and impact parameter b = 1.00×10¹¹ m; GM_☉ = 1.327×10²⁰ m³s⁻².

Unbound: E > 0, expect e > 1. A

E = ½mv_∞², L = mv_∞b, k = GM_☉m

Angular momentum from the far-field: perpendicular distance b times momentum. B

e = √(1 + 2EL²/mk²) = √(1 + v_∞⁴b²/(GM_☉)²) = √(1 + (v_∞²b/GM_☉)²)

The mass cancels; introduce the dimensionless group v_∞²b/GM_☉. C

v_∞²b/GM_☉ = (6.76×10⁸)(1.00×10¹¹)/1.327×10²⁰ = 0.5094

Evaluate the group. A

e = √(1 + 0.2595) = 1.122; p = v_∞²b²/GM_☉ = 5.09×10¹⁰ m; r_min = p/(1+e)

Perihelion of a conic is at θ = 0. B

e ≈ 1.12, r_min = 5.09×10¹⁰/2.122 = 2.40×10¹⁰ m ≈ 0.16 AU

Reading. A genuine hyperbolic pass (e > 1): the body swings to within 0.16 AU of the Sun and leaves forever. The turn (deflection) angle between the incoming and outgoing asymptotes is Φ = 2 arcsin(1/e) = 2 arcsin(0.891) ≈ 126°. Units check. v_∞²b/GM = (m²s⁻²)(m)/(m³s⁻²) = dimensionless; r_min in metres.

Problems

A satellite in a bound orbit has perihelion r_p = 6.6×10⁶ m and aphelion r_a = 4.2×10⁷ m about Earth (GM_⊕ = 3.986×10¹⁴). Find a and e.
Solution
a = (r_p + r_a)/2 = (6.6×10⁶ + 4.2×10⁷)/2 = 2.43×10⁷ m. e = (r_a − r_p)/(r_a + r_p) = (3.54×10⁷)/(4.86×10⁷) = 0.728. This is a typical geostationary-transfer ellipse.
For the orbit of Problem 1 (satellite mass m = 1200 kg), compute the total energy E and the magnitude of the LRL vector A.
Solution
k = GM_⊕m = 3.986×10¹⁴×1200 = 4.783×10¹⁷. E = −k/2a = −4.783×10¹⁷/(2×2.43×10⁷) = −9.84×10⁹ J. The LRL magnitude is |A| = mke = 1200×4.783×10¹⁷×0.728 = 4.18×10²⁰ kg² m³ s⁻². Check: A/mk = 4.18×10²⁰/(1200×4.783×10¹⁷) = 0.728 = e. ✓
Show that for a circular orbit the LRL vector vanishes, and interpret the result physically.
Solution
A circular orbit has e = 0. Since |A| = mke, A = 0. Physically, A points toward perihelion; a circle has no unique perihelion (every point is equidistant), so no preferred direction can exist and A must be zero. Equivalently, p×L has magnitude pL = m²v²r and for a circle mv²/r = k/r² ⇒ m²v²r = mk, which exactly cancels the mkr̂ term.
A comet is observed with eccentricity e = 0.999 and perihelion distance r_p = 0.60 AU. Compute its semi-major axis and comment on its orbital period. (1 AU = 1.496×10¹¹ m, GM_☉ = 1.327×10²⁰.)
Solution
r_p = a(1−e) ⇒ a = r_p/(1−e) = 0.60/(0.001) = 600 AU = 8.98×10¹³ m. Period (Kepler III, T = 2π√(a³/GM_☉)): a³ = 7.24×10⁴¹, /GM = 5.46×10²¹, √ = 7.39×10¹⁰ s, ×2π = 4.64×10¹¹ s ≈ 1.47×10⁴ yr. A long-period comet, nearly parabolic but still bound.
Starting from dA/dt = 0, prove that A lies in the orbital plane and points along the major axis, and show |A| = mke using A·r.
Solution
(i) A·L = (p×L)·L − mkr̂·L. The first term vanishes ((p×L)⊥L); the second vanishes because r (hence r̂) is perpendicular to L = r×p. So A·L = 0: A lies in the orbital plane. (ii) From A·r = L² − mkr = Arcosθ, solving gives r = (L²/mk)/(1 + (A/mk)cosθ), a conic with perihelion at θ = 0, i.e. along A; hence A points to perihelion (the major axis). Comparing with r = p/(1+ecosθ) gives A/mk = e, so |A| = mke. ■